Water is dripping out at a steady rate of $1 \text{ cm}^3/\text{sec}$ through a tiny hole at the vertex of a conical vessel,whose axis is vertical. When the slant height of water in the vessel is $4 \text{ cm}$,find the rate of decrease of the slant height,where the semi-vertical angle of the conical vessel is $\frac{\pi}{6}$.

(N/A) Let $v$ be the volume of water in the conical vessel at time $t$. We are given that $\frac{dv}{dt} = -1 \text{ cm}^3/\text{s}$ (negative because water is dripping out).
Let $l$ be the slant height,$h$ be the vertical height,and $r$ be the radius of the water surface.
The semi-vertical angle is $\alpha = \frac{\pi}{6}$.
From the geometry of the cone,we have $h = l \cos \alpha = l \cos \frac{\pi}{6} = l \frac{\sqrt{3}}{2}$ and $r = l \sin \alpha = l \sin \frac{\pi}{6} = \frac{l}{2}$.
The volume of the cone is $v = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{l}{2}\right)^2 \left(l \frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3} \pi}{24} l^3$.
Differentiating with respect to $t$,we get $\frac{dv}{dt} = \frac{\sqrt{3} \pi}{24} \cdot 3l^2 \frac{dl}{dt} = \frac{\sqrt{3} \pi}{8} l^2 \frac{dl}{dt}$.
Given $l = 4 \text{ cm}$ and $\frac{dv}{dt} = -1 \text{ cm}^3/\text{s}$,we substitute these values:
$-1 = \frac{\sqrt{3} \pi}{8} (4)^2 \frac{dl}{dt} = \frac{\sqrt{3} \pi}{8} \cdot 16 \frac{dl}{dt} = 2\sqrt{3} \pi \frac{dl}{dt}$.
Thus,$\frac{dl}{dt} = -\frac{1}{2\sqrt{3} \pi} \text{ cm/s}$.
The rate of decrease of the slant height is $\frac{1}{2\sqrt{3} \pi} \text{ cm/s}$.